Okay, I have 2 questions that I can't figure out...I've manipulated them in so many ways, but it just doesn't seem like i'm doing it right, can someone please show me how they should be answered.

1. Frank and Gordon have decided to go on seperate vacations. At 9am this morning Gordon's plane was exactly 300km south of Frank's plane. Frank's plane was flying west at the rate of 650 km/h and Gordon's plane is flying north at a rate of 525km/h
a. At what rate is the distance between Frank and Gordon changing at 9:15am?

b. When do Frank and Gordon's planes cease to approach one another?

2. After his vacation, Gordon decided he needed to get into shape. Frank agreed to build him a 400m track in the shape of two semi-circles at the ends of a rectangle. The straight sections of the track must be at least 100m in length, and the radius of the semi-circle must be at least 20m. Find the dimensions of the track that enclose the maximum area and the minimum area.

2. Originally Posted by sugar_babee
...
1. Frank and Gordon have decided to go on seperate vacations. At 9am this morning Gordon's plane was exactly 300km south of Frank's plane. Frank's plane was flying west at the rate of 650 km/h and Gordon's plane is flying north at a rate of 525km/h
a. At what rate is the distance between Frank and Gordon changing at 9:15am?

b. When do Frank and Gordon's planes cease to approach one another?
...
Hello,

I'm going to use a coordinate system: Frank's plane starts in (0,0) and Gordon's plane starts in (0, -300). The y-axis is pointing North and the x-axis is pointing East. The variable t represents the elapsed time and is measured in hours.

The course of Frank's plane on the x-axis is: $f(t)=-650\cdot t$
The course of Gordon's plane on the y-axis is: $g(t)=-300+525\cdot t$

The distance between both planes is calculated using the formula for the distance between two points:

$d(t)=\sqrt{(0-(-650\cdot t))^2+(-300+525\cdot t-0)^2}$ = $\sqrt{698125t^2-315000 t+90000}$

The rate of change is d'(t):

$d'(t)=\frac{25(1117 t-252)}{\sqrt{1117t^2-504 t+144}}$

to a) At 9:15 the elapsed time is t=0.25 hours. Plug in this value into d'(t) and you'll get the rate of change: $d'\left(\frac{1}{4}\right)\approx 72,7 \frac{km}{h}$

to b) The approach will cease if the rate of change equals zero:
$d'(t)=0 \text{ if } 1117 t-252=0 \Longleftrightarrow t= \frac{252}{1117}
\approx 13 \text{min}\ 32\text{s}$
when the planes are 233.38 km apart(is that the right expression?)

EB

3. Originally Posted by earboth
Hello,

I'm going to use a coordinate system: Frank's plane starts in (0,0) and Gordon's plane starts in (0, -300). The y-axis is pointing North and the x-axis is pointing East. The variable t represents the elapsed time and is measured in hours.

The course of Frank's plane on the x-axis is: $f(t)=-650\cdot t$
The course of Gordon's plane on the y-axis is: $g(t)=-300+525\cdot t$

The distance between both planes is calculated using the formula for the distance between two points:

$d(t)=\sqrt{(0-(-650\cdot t))^2+(-300+525\cdot t-0)^2}$ = $\sqrt{698125t^2-315000 t+90000}$

The rate of change is d'(t):

$d'(t)=\frac{25(1117 t-252)}{\sqrt{1117t^2-504 t+144}}$

to a) At 9:15 the elapsed time is t=0.25 hours. Plug in this value into d'(t) and you'll get the rate of change: $d'\left(\frac{1}{4}\right)\approx 72,7 \frac{km}{h}$

to b) The approach will cease if the rate of change equals zero:
$d'(t)=0 \text{ if } 1117 t-252=0 \Longleftrightarrow t= \frac{252}{1117}
\approx 13 \text{min}\ 32\text{s}$
when the planes are 233.38 km apart(is that the right expression?)

EB
I did it a different way and got a different answer. I suspect I've done it the wrong way though. (Highest level of Mathematic education = Grade 10, which I'm in right now). I thought I'd give it a go and see if I could help.

My answers (I'll post my method in another post) were:

a) The rate at which the distance between Frank and Gordon is changing is 79.16km/h.

b) The planes cease to approach each other when Gordon reaches Franks starting place, 300km North. The time this happens is 9:34 am.

Hope that helps (oh, and take in mind that they're rounded).

4. Originally Posted by Kelsey-Lee
I did it a different way and got a different answer. I suspect I've done it the wrong way though. (Highest level of Mathematic education = Grade 10, which I'm in right now). I thought I'd give it a go and see if I could help.

My answers (I'll post my method in another post) were:

a) The rate at which the distance between Frank and Gordon is changing is 79.16km/h.

b) The planes cease to approach each other when Gordon reaches Franks starting place, 300km North. The time this happens is 9:34 am.

Hope that helps (oh, and take in mind that they're rounded).
Kelsey-Lee: I'm tending toward Earboth's solution as being correct (I would do it his way and I agree with his numbers), but I can't check yours until I know how you did it. Can you post your method?

-Dan

5. Thanks a lot, I see what i did wrong now...Any suggestions on the second question?

2. After his vacation, Gordon decided he needed to get into shape. Frank agreed to build him a 400m track in the shape of two semi-circles at the ends of a rectangle. The straight sections of the track must be at least 100m in length, and the radius of the semi-circle must be at least 20m. Find the dimensions of the track that enclose the maximum area and the minimum area.

I somehow ended up with minimum area of 0...that cant be right

6. Originally Posted by sugar_babee
Thanks a lot, I see what i did wrong now...Any suggestions on the second question?

2. After his vacation, Gordon decided he needed to get into shape. Frank agreed to build him a 400m track in the shape of two semi-circles at the ends of a rectangle. The straight sections of the track must be at least 100m in length, and the radius of the semi-circle must be at least 20m. Find the dimensions of the track that enclose the maximum area and the minimum area.

I somehow ended up with minimum area of 0...that cant be right
(All lengths in metres)

So we have $s \ge 100$, and $r \ge 20$, where $s$ is the length of the straight section, and $r$ the radius of the semi-circles.

The length $l$ of the track is $400$, so:

$
l=2s+2 \pi r=400
$

so:

$r=(200-s)/\pi$

and as $s \ge 100$, this implies that $r \le 100/\pi$, that is:

$r \in [20,100/\pi]$.

Now the area $A$ enclosed by the track is:

$A=2.r.s+\pi\ r^2$

but $s=200-\pi r$, so:

$
A=2.r[200-\pi r]+\pi r^2=400r-\pi r^2
$

Now a differentiable function on a closed interval takes its maximum and minimum values either at a stationary point of the function, or at the end points of the interval. So lets look for the stationary points of $A$:

$\frac{dA}{dr}=400-2 \pi r$, so the only stationary point of $A$ is at the solution of:

$400-2 \pi r=0$,

or:

$r=200/\pi$.

But this stationary point is not in the interval in which $r$ must lie $r \in [20,100/\pi]$.

So the area takes its maximum and minimum values at the end points of the interval $[20,100/\pi]$ for $r$. Evaluate it at these points and the two values will be the maximum and minimum areas (in some order).

RonL